Basic trigonometry in dynamics force problem

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Homework Help Overview

The problem involves two spheres with equal mass and electric charge, suspended by threads and forming an angle due to the repulsive force between them. The objective is to determine the value of the charge based on the geometry and forces involved.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to use trigonometric relationships and force balance to derive the charge, expressing confusion about different methods encountered online. They question the reasoning behind certain conclusions related to trigonometric relations.

Discussion Status

The discussion includes attempts to clarify the original poster's understanding of vectors and trigonometric applications in the context of the problem. Some participants offer brief affirmations of help, but there is no detailed resolution or consensus on the questions raised.

Contextual Notes

The original poster expresses uncertainty about their grasp of trigonometric relations and vectors, indicating a potential gap in foundational understanding that may affect their approach to the problem.

Bunny-chan
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Homework Statement


Two spheres, with 0.5g each, are hanging by 30cm threads, tied on the same spot. The same electric charge is communicated to each sphere; in consequence, the threads move apart until they are about 60^\circ from each other. What is the value of the charge?

\theta = \frac{60^\circ}{2} = 30^\circ \\<br /> \\m = 0.5g = 0.0005kg\\ d = 30cm = 0.3m


Homework Equations


Not needed.

The Attempt at a Solution


So, what I did was the following:

\tan \theta = \frac{\vec F}{\vec P} \\ \Rightarrow \frac{\sqrt 3}{3} = \frac{\vec F}{m\vec g} = \frac{\vec F}{0.0005 \times 9.8} = \frac{\vec F}{0.005} \\ \Rightarrow \vec F = \frac{\sqrt 3}{3} \times 0.005 = 0.0028 \\k \frac{q^2}{d^2} = 0.0028 \\ \Rightarrow 9 \times 10^9 \times \frac{q^2}{0.3^2} = 2.8 \times 10^{-3} \\ \Rightarrow q^2 = \frac{(2.8 \times 10^{-3}) \times (9 \times 10^{-2})}{9 \times 10^9} = 2.8 \times 10^{-14} \\ \Rightarrow q = 1.7 \times 10^{-7}CMy result did match the answer in the textbook, but searching through the web, I came across a different way of solving it:

v6h0Lxr.png

The rest is just like mine, so I didn't bother putting it. Anyway, he reached the same result, but with a different method and trigonometric relations, which I'm quite lacking in and I couldn't understand very well yet.

Why did he conclude that
ca416d5dfd214713807a96b6c4bfef7d.png


and
56f040f51842463aa3cf040d13f2222e.png

?

I know I could just go on and ignore this because I already solved the exercise, but vectors are a puzzling topic for me and I feel like there's something I'm missing... I'd appreciate some help!
 
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i hope that help
prove.jpg
 
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the d part :
prove 2.jpg
 
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